Rouche's Theorem for $p(z)=z^5+10z-3$ > Let $f$ and $g$ be differentiable on a domain $D$ and suppose that $\gamma$ is a simple closed contour whose inside is contained in D. > > If $|f(z)-g(z)|<|f(z)|$ for all $z$ on $\gamma$, then $f$ and $g$ have the same number of zeros inside $\gamma$ (counted including their order). I was reading an example of application of Rouche's Theorem, where Rouche's theorem was used to show that the polynomial $p(z)=z^5+10z-3$ has $1$ zero in $\\{z:\mathbb{C}:|z|<1\\}$. What was done: Let $f(z)=10z-3$ and $g(z)=z^5+10z-3$. Let $|z|=1$. Then we get that: $|f(z)-g(z)|\\\=|-z^5|= |z^5|=1<z\le|10z-3|$ Hence $p(z)$ has 1 zero inside $\\{z:\mathbb{C}:|z|<1\\}$. * * * How did they get that $1<z$? What actually is $z$? Secondly, why did they choose $f(z)=10z-3$? Can I choose $f(z)=10z$ and is what I did below correct? $|f(z)-g(z)|\le|f(z)|\\\|-z^5+3|\le10|z|\\\|z|^5+|3|\le10|z|\\\4\le10$ Hence p(z) has 1 zero inside $\\{z:\mathbb{C}:|z|<1\\}$.

So the only thing that concerns me about your formulation is that you assumed $|f(z)-g(z)|\leq |f(z)|$, rather than showing it. If we try to work backwards, starting with $4<10$ (the inequality needs to be strict), then we can recover $|-z^5| + |3| < 10|z|$ without a problem. Then we can use the triangle inequality because $|-z^5 + 3|\leq |-z^5|+|3|$, which recovers the step above. From there, we just fill in the missing terms to get the desired result $|f(z)-g(z)| < |f(z)|$.

Therefore, you can apply Rouché's Theorem. So you did it "right" but backwards. The reason they chose $10z-3$ instead is probably just so they didn't have to deal with the triangle inequality.

Edit: And yes, I agree with Adeal about $|z|$.